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\textbf{Mathematics 468} \\
\textbf{Homework 3 solutions}
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Given a subset $A$ of a topological space $X$, let $A'$ be the set of
limit points of $A$.  Let $\closure$ denote the closure of $A$: let
$\closure = A \cup A'$.

1. Show that $\closure$ is closed.

\textsc{Solution:} By a result from class, if I can show that
$\closure$ contains all of its limit points, then I can conclude that
$\closure$ is closed.  Let $y$ be a limit point of $\closure$.  I
claim that every neighborhood of $y$ intersects $A$, so that $y$ is
contained either in $A$ or in $A'$.  If $y \in A$, then there is
nothing more to show, so I'll assume that $y \not \in A$, and I'll
show that every neighborhood of $y$ intersects $A$.  Let $N$ be a
neighborhood of $y$.  If $N$ contains a point of $A$, then that's
good.  Otherwise, $N$ must contain a point $x$ of $A'$.  Since $x$ is
a limit point of $A$, then every neighborhood of $x$ must contain a
point of $A$; in particular, $N$ is a neighborhood of $x$, and so must
contain a point of $A$.  Hence every neighborhood $N$ of $y$ contains
a point of $A$, and hence $y$ is a limit point of $A$, and hence $y$
is in $\closure$.  Therefore $\closure$ contains all of its limit
points, so it is closed.

2. Show that $\closure$ is the smallest closed subset of $X$
containing $A$.  In other words, if $C$ is a closed set containing
$A$, show that $C$ contains $\closure$ also.  (Equivalently, show that
$\closure$ is the intersection of all closed sets containing $A$.)

\textsc{Solution:} Suppose that $C$ is a closed set containing $A$.  I
want to show that $C$ contains $\closure = A \cup A'$; in other words,
I want to show that $C$ contains both $A$ and $A'$.  By assumption,
$C$ contains $A$, so given $y$ in $A'$ but not in $A$, I want to show
that $y$ is in $C$.  Every neighborhood of $y$ intersects $A$ in some
point other than $y$, and since $A \subseteq C$, then every
neighborhood of $y$ intersects $C$ in some point other than $y$.
Since $C$ is closed, then $C$ contains all of its limit points, so $y$
must be contained in $C$.

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